There's nothing more relaxing than a nice glass of lemonade on a hot summer day. But, did you know you are actually doing chemistry when you make it? The amount of lemonade powder you put into the glass, combined with the amount of water you put in to make the perfect concentration is molarity in action!
- This article covers molarity.
- First, we will define molarity and learn it's related equation.
- Next, we will learn how to find moles in molarity related problems.
- Thereafter, we will cover how to calculate the molarity of a diluted solution.
- Lastly, we will learn how to calculate the molarity of a mixed solution.
Definition of Molarity
Let's start by looking at the definition of molarity.
Molarity is the concentration of solute dissolved in a solution expressed in units of moles per liter.
Molarity, or molar concentration, describes the concentration of an amount of a substance dissolved in a liquid. We call the substance we are dissolving a solute and the liquid is called a solvent. Specifically, molarity is defined by the number of moles per liter: mol/L.
Solutes can consist of anything that dissolves into a liquid; they can be solids, other liquids, or even gases. If you know the amount of a solute in moles and the volume of solvent it is dissolved into, finding molarity is simple!
Molarity Equation
The standard molarity equation is thankfully very simple! It is :
$$Molarity\,(M)=\frac{n_{solute}}{V_{solution}}$$
The three variables are defined as:
M is molar concentration expressed in mol/L
n is the molar amount of the solute expressed in mol
V is the volume of the solution expressed in L
How to find moles in molarity problems
Often, molarity problems won't just be as simple as dividing the moles of the solute by the liters of the solution. It is only one step in more complex problems. The beginning steps can involve many different things, but they will all lead to finally finding the amount of solute in moles and the volume in liters!
Instead of a problem just giving you moles, it may give you the number of total particles of the solute, the mass of solute used, or a reaction that creates the solute.
Let's take a look at a problem: it may seem complicated, but remember your end goal - you only need to find the total amount of moles of solute and the total volume of solution.
A student is preparing a nice bowl of soup, find the molarity of salt (NaCl) if this is the recipe:
1.5 liters of Water
60 grams of Salt
0.5 kg of Pasta
0.75 liters of Chicken Stock
200 grams of salted butter (3% salt by weight)
- Isolate the sources of solute aka. salt:60g of Salt (100% Salt)200 grams of salted butter (3% salt)
- Find the molar mass of solute, which is salt in this example: $$Na\,(22.98\frac{g}{mol})+Cl\,(35.45\frac{g}{mol})=58.44\frac{g}{mol}$$
- Calculate moles of solute (salt) in pure salt: $$\frac{60\,g}{58.44\frac{g}{mol}}=1.027\,mol$$
- Find the weight of salt in butter: $$200\,g*3\%=6\,g\,NaCl$$
- Calculate moles of salt in butter: $$\frac{6\,g}{58.44\frac{g}{mol}}=0.1027\,mol$$
- Add both sources of salt to find total moles: $$1.027\,mol+0.1027\,mol=1.129\,mol$$
- Total all solvents used: $$1.5\,L+0.75\,L=2.25\,L\,H_2O$$1.5l+0.75l=2.25l of water
- Divide moles of solute by liters of solvent: $$\frac{1.129\,mol}{2.25\,L}=0.501\,M$$
Even though this problem was a lot of steps, as long as you keep your end goal in mind it is easy to work towards the solution! Always remember you need to find the total amount of solute and the total volume of solution.
If you run into any trouble following any of these steps, it might help to refresh your knowledge on moles and molar mass in general.
Uses of Molarity
When reacting chemicals you almost always use solutions. In general, it is very hard to react two dry chemicals so one or both of your reactants must be in a solution. Just as it is with any chemical reaction moles are the key players, even if the reaction takes place in solution.
So, you probably need to calculate the mole ratios too. Luckily, these mole ratios don't even have to be calculated with moles, they can be directly calculated with molarity. Since molarity is always expressed with respect to a single liter, the mole ratio stays the same.
If you have the molarity of a solution and the volume of the solution it is very easy to calculate the moles in that solution. Just multiply both sides of the molarity equation by the volume to give you:
$$M_1V_1=n_1$$
Let's use this equation in a simple precipitation reaction with two solutions
$$Pb(NO_3)_{2\,(aq)} + 2KI_{(aq)} \rightarrow 2KNO_{3\,(aq)} + PbI_{2\,(s)}$$
Using this reaction, find the volume of 1.2M KI(aq) solution required to create 1.5 moles of PbI2 if reacted with excess amounts of Pb(NO3)2(aq).
- Find the mole ratio of KI to PbI2:2 KI to make 1 PbI2
- Calculate the amount of KI needed: $$1.5\,mol,PbI_2*\frac{2\,mol\,KI}{1\,mol\,PbI_2}=3\,mol\,KI$$
- Calculate volume of solution needed: $$\frac{3\,mol}{1.2\frac{mol}{L}}=2.5\,L\,KI_{(aq)}$$
This problem is a simple example of how molarity is used in real chemical reactions. It is a critical component of almost every reaction
How to calculate dilutions using molarity
If you ever have to make a solution in the lab, or just want to pass your AP Chemistry exam, you are going to need to get used to molarities. One of the best uses of molarity is to calculate dilutions fast! In the lab, we usually only have a couple of solutions that are created at specific molarities. These solutions are called stock solutions.
A stock solution is a standardized solution of precisely known molar concentration that will be found in labs in large volumes
A stock solution of 2.0 M hydrochloric acid (HCl) is easy to produce and can be stored for a long time. Usually, however, you would need lower concentrations of HCl, think like 0.1 M or so, to do your reaction. In order to create this lower concentration solution, you must dilute the stock solution by adding more solvent. In some experiments such as titrations, low concentration acids and bases are more effective as they are easier to control. Thankfully there is an easy way to calculate the needed dilutions, just use this equation:
$$M_1V_2=M_2V_2$$
M1 & V1 refer to the volume and molarity of the stock solution, respectively. Usually, you will leave V1 as a variable as you are trying to find the volume of the solution you will need. V2 & M2 refer to the molarity and volume of the solution you are trying to make. Let's see an example to show how it would work in a lab:
When performing experiments, an independent variable will always have to change. Testing over a wide range of concentrations of a solution can show if the concentration has an impact on the dependent variable.
For an experiment, you want to test if the concentration of salt in water affects its ability to conduct electricity. To test this, you want to create solutions with molarities of 5M and 1M, each having 2L total. First, create a solution of 5M NaCl with solid salt, then create the 1M solution by diluting the 5M solution.
First, create the 5M solution,
Find the amount of salt in grams needed
Moles of salt will be \(5\,M*2\,L=10\,mol\)
For the mass of salt: $$58.55\frac{g}{mol}*10\,mol=585.5\,g$$
Add this amount of salt to 2L of water, resulting in the 5M solution.
Second, dilute the 5M solution to create 2L of 1M solution
$$M_1V_2=M_2V_2$$
$$5\,M(V_1)=1\,M(2\,L)$$
$$V_1=\frac{1\,M*2\,L}{5\,M}=0.4\,L$$
Add 0.4L of the 5M to a beaker, then add enough water for the total volume to equal 2L. This means you will only have to add 1.6L of water. Remember, it is the total volume that needs to be 2L, not the amount of water you add.
So, to recap:
the first solution will need 585.5g of salt and 2L of water
the second solution will need 0.4L of the 5M solution and 1.6L of water
Molarity of Multiple Solutions Mixed
Sometimes you may end up having to find the concentration of two solutions after mixing them. It might seem complicated, but remember the steps to the original problem solving: 1st- find the total moles & 2nd- find the total volume!
Suppose you have multiple solutions with multiple volumes. You have to store this solution long term, but you only have one appropriate container for all of it. You decide to mix them all together but need to figure out the total volume and final molarity of it all.
Solution 1 is 3.0M and you have 0.5L of it.
Solution 2 is 1.5M and you have 0.75L of it
and Solution 3 is 0.75M and you have 1.0L of it
Find the final molarity after mixing all three solutions.
To start, you want to find the total moles present of solute that will be in the final mixture.
This is easily accomplished by adding up the moles of solute in each solution.
For Solution 1, this will be \(M_1V_1=n_1\): $$3.0\,M(0.5\,L)=1.5\,mol$$
For Solution 2, this will be \(M_2V_2=n_2\): $$1.5\,M(0.75\,L)=1.125\,mol$$
For Solution 3, this will be \(M_3V_3=n_3\): $$0.75\,M(1.0\,L)=0.75\,mol$$
For the total it will be \(n_1+n_2+n_3\): $$1.5\,mol+1.125\,mol+0.75\,mol=3.375\,mol$$
Now, find the total volume which will be \(V_1+V_2+V_3\): $$0.5\,L+0.75\,L+1.0\,L=2.25\,L$$
Finally, like before, divide total moles by total volume: $$\frac{3.375\,mol}{2.25\,L}=1.5\,M$$
So from the example, it is easy to see what the equation should be when mixing any amount of solutions with the same solute. Divide the total moles by the total volume!
Total moles in the solution will be \(n_1+n_2+n_3+...,\), but this will be \(M_1V_1+M_2V_2+M_3V_3+...,\)
Total volume is simply \(V_1+V_2+V_3+...,\)
Dividing these leaves you with:
$$M_{solution}=\frac{M_1V_1+M_2V_2+...,}{V_1+V_2+...,}$$
Molarity - Key takeaways
- Molarity is the concentration of solute dissolved in a solution expressed in units of moles per liter
- The standard molarity equation is: $$Molarity\,(M)=\frac{n_{solute}}{V_{solution}}$$
M is molar concentration expressed in mol/L
n is the molar amount of the solute expressed in mol
V is the volume of the solution expressed in L
A stock solution is a standardized solution of precisely known molar concentration that will be found in labs in large volumes
To find the new molarity for dilutions, use the following equation: $$M_1V_2=M_2V_2$$
The total molarity of a solution is: $$M_{solution}=\frac{M_1V_1+M_2V_2+...,}{V_1+V_2+...,}$$
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